Question:

In amplitude modulation, if the modulation index is \(0.8\) and the peak voltage of the message signal is \(10\,\text{V}\), then the amplitudes of the upper and lower side bands of the modulated wave are respectively

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For amplitude modulation, \[ m=\frac{V_m}{V_c} \] and the amplitude of each sideband is \[ V_{SB}=\frac{mV_c}{2}. \] The upper and lower sidebands always have equal amplitudes.
Updated On: Jul 9, 2026
  • \(22.5\,\text{V},\;2.5\,\text{V}\)
  • \(12.5\,\text{V},\;10\,\text{V}\)
  • \(5\,\text{V},\;5\,\text{V}\)
  • \(10\,\text{V},\;5\,\text{V}\) \bigskip
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The Correct Option is C

Solution and Explanation

Concept: For an AM wave, \[ m=\frac{V_m}{V_c}, \] where \[ m=\text{modulation index}, \quad V_m=\text{peak message voltage}, \quad V_c=\text{carrier peak voltage}. \] The amplitude of each sideband is \[ V_{SB}=\frac{mV_c}{2}. \] The upper and lower sidebands have equal amplitudes.

Step 1:
Calculate the carrier amplitude. Given, \[ m=0.8, \qquad V_m=10\,\text{V}. \] Using \[ m=\frac{V_m}{V_c}, \] \[ 0.8=\frac{10}{V_c}. \] \[ V_c=\frac{10}{0.8}. \] \[ V_c=12.5\,\text{V}. \]

Step 2:
Find the amplitude of each sideband. \[ V_{SB} = \frac{mV_c}{2}. \] \[ V_{SB} = \frac{0.8\times12.5}{2}. \] \[ V_{SB} = \frac{10}{2}. \] \[ V_{SB}=5\,\text{V}. \]

Step 3:
Write the amplitudes of USB and LSB. \[ V_{USB}=5\,\text{V}, \] \[ V_{LSB}=5\,\text{V}. \]

Step 4:
Write the final answer. \[ \boxed{V_{USB}=5\,\text{V},\quad V_{LSB}=5\,\text{V}} \] \[ \boxed{\text{Answer = (C)}} \]
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